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Now, I am aware of the summation convention for the dot product

$$\mathbf{a} \cdot \mathbf{b} = a_i b_i$$

But I am unsure about how to represent $(\nabla \cdot \mathbf{a}) \mathbf{b}$ and $(\mathbf{a} \cdot \nabla) \mathbf{b}$, since in general, $$(\nabla \cdot \mathbf{a}) \mathbf{b} \ne (\mathbf{a} \cdot \nabla) \mathbf{b}$$

In either case, I come up with $\nabla_i a_i b_j$ = $ a_i \nabla_i b_j$, since (from what I understand) the order of the terms ought not to matter. This is ambiguous, though, as it doesn't differ between the two. Can anyone hint me in the correct direction?

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The order of the terms can matter. This is especially true of derivative operators and the objects on which they operate. For example,

$$\begin{align} (\nabla \cdot \vec a)\vec b&=(\partial_i a_i)b_j\\\\ &=b_j(\partial_i a_i)\\\\ &\ne a_i\partial_i(b_j) \,\,\text{(and certainly not}\,\,a_ib_j\partial_i)\\\\ &=(\vec a\cdot \nabla)\vec b \end{align}$$

I have actually found retaining unit vectors where applicable can help. In the previous example, I might have written

$$(\nabla \cdot \vec a)\vec b=(\partial_i a_i)(\hat x_jb_j)$$

For more complex expressions, this can be quite useful. For example, suppose we have $\vec A \times (\vec B\times \vec C)$. Then, I would write

$$\begin{align} \vec A \times (\vec B\times \vec C)&=\hat x_i A_i\times (\hat x_j B_j\times \hat x_kC_k)\\\\ &=(\hat x_i\times (\hat x_j\times \hat x_k))A_iB_jC_k\\\\ &=(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k)A_iB_jC_k\\\\ &=(A_iC_i)\hat x_jB_j-(A_iB_i)\hat x_kC_k\\\\ &=(\vec A\cdot \vec C)\vec B-(\vec A\cdot \vec B)\vec C \end{align}$$

recovering the familiar vector triple product!

And in one last example, from THIS ANSWER I POSTED HERE

$$\begin{align} \nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\ &=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\ &=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\ &=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\ &=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\ &=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5 \end{align}$$

In going from $(1)$ to $(2)$ we made use of the vector triple product. Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.

In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.

In going from $(3)$ to $(4)$, we rearranged terms.

In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.

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